Problem: What's the first wrong statement in the proof below that $ \triangle ABC \cong \triangle EBC$ $ \; ?$ $ \overline{BC} $ is parallel to $ \overline{DF} $. This diagram is not drawn to scale. A B C D E F Givens $ \overline{DE} \cong \overline{AC}$ $, \ $ $ \angle BDE \cong \angle ACB$ $, \ $ $ \angle DBE \cong \angle ABC$ $, \ $ $ \overline{CF} \cong \overline{BC}$ $, \ $ $ \angle CFE \cong \angle ABC$ $, \ $ and $\ $ $ \angle CEF \cong \angle BAC$ Proof $ \triangle ABC \cong \triangle EBD$ because AAS $ \overline{AB} \cong \overline{BE}$ because corresponding parts of congruent triangles are congruent $ \angle CBE \cong \angle DBE$ because alternate interior angles are equal $ \triangle ABC \cong \triangle EFC$ because AAS $ \overline{EF} \cong \overline{AB}$ because corresponding parts of congruent triangles are congruent $ \overline{AC} \cong \overline{CE}$ because corresponding parts of congruent triangles are congruent $ \triangle ABC \cong \triangle EBC$ because SSS
Explanation: Try going through the proof yourself: write down the givens, and then see if they justify the next step for the reason given. Then do the same thing for the next step, and the next, until you run into something that you can't justify, or you finish the proof. $ \angle DBE \cong \angle CBE$ is the first wrong statement.